1
vote
1answer
61 views

Complement but not direct summand; Lam, Lectures on Modules and Rings, Example 6.17(5)

Let $S$ be a submodule of an $R$-module $M$. A submodule $C⊆M$ is said to be a complement to $S$ (in $M$) if $C$ is maximal with respect to the property that $C∩S=0$. (This does exist by Zorn Lemma.) ...
3
votes
0answers
44 views

Commutativity of direct and inverse limits

In exercise 5.34(iv) of Homological Algebra book by Rotman one is asked to prove that direct limits and inverse limits do not necessarily commute. I have two questions : 1.) Is it true that ...
1
vote
2answers
39 views

A finite $\mathbb{Z}$-module whose submodules are totally ordered by inclusion.

I have the following problem: Let $M$ be a finite $\mathbb{Z}$-module such that set of the submodules is totally ordered by inclusion. Prove that there exist a prime $p$ such that $|M|=p^\alpha$ ...
0
votes
0answers
31 views

A proposition on exact sequence of inverse limit (Lang, Algebra, p. 165)

I am trying to understand this proof. My only question is that what are the vertical maps here?
3
votes
1answer
55 views

Classify abelian groups $A$ which are irreducible $End(A)$-modules

Classify abelian groups $A$ which are irreducible $End(A)$-modules. I think i did it for finite abelian group $A$ . A finite abelian group $A$ is irreducible iff order of $A$ a is power of prime. ...
0
votes
0answers
30 views

Identifying the abelian group with a presentation matrix

I am doing problems from Artin: \begin{bmatrix} 2 \\ 1\\ \end{bmatrix} and \begin{bmatrix} 2 & 4\\ 1 & 4\\ \end{bmatrix} For the First one after manipulating rows I ...
17
votes
2answers
337 views

If the tensor power $M^{\otimes n} = 0$, is it possible that $M^{\otimes n-1}$ is nonzero?

Let $M$ be a module over a commutative ring $R$. It is possible that $M \otimes M = 0$ if $M$ is nonzero, for example when $R = \mathbb{Z}$ and $M = \mathbb{Q}/ \mathbb{Z}$. What about when higher ...
0
votes
2answers
116 views

Possible difference between $\mathbb{Z}$-modules and vector spaces

Suppose $G$ is a free abelian groups, i.e. a free $\mathbb{Z}$-module; we have a set $S \subset G $ such that $S$ spans $G$. Can we conclude that the rank of $G$ as a $\mathbb{Z}$-module is $ \leq ...
3
votes
1answer
78 views

Maximal $\mathbb{Z}$-submodules in $\mathbb{Q}$

Is it true that $\mathbb{Q}$ viewed as $\mathbb{Z}$-module ( i.e. abelian group ) has not maximal $\mathbb{Z}$-submodules ? Why ?
7
votes
2answers
118 views

Is $\Bbb Q/\Bbb Z$ artinian as a $\Bbb Z$-module?

I'm confused. Is $\Bbb Q/\Bbb Z$ artinian as a $\Bbb Z$-module? We know that $\Bbb Z_{p^{\infty}} \subset \Bbb Q/\Bbb Z$ is artinian. The following argument is true or not ? $\mathbb Q / ...
2
votes
1answer
146 views

Classify Artinian $\Bbb Z$-modules

How can I classify Artinian $\mathbb{Z}$-modules as Noetherian $\mathbb{Z}$-module? (A $\mathbb{Z}$-module is Noetherian iff it is finitely generated). Any hint will be helpful. I have seen the ...
1
vote
0answers
129 views

Isomorphism theorem for Abelian groups, related to Hatcher exercise 2.1.14

I am trying to understand which Abelian groups can fit the short exact sequence \begin{equation} 0 \rightarrow \mathbb{Z}_{p ^m}\rightarrow A \rightarrow \mathbb{Z}_{p^n}\rightarrow 0. \end{equation} ...
0
votes
1answer
72 views

Why $\mathbb Z_{p^{\infty}}$ is an artinian $\mathbb Z$-module? [duplicate]

If $M=\{\frac{a}{p^{n}}| a\in\mathbb Z , n\in\mathbb N\}$ and quotient group $(M/\mathbb Z,+)=\mathbb Z_{p^{\infty}}$. Prove that $\mathbb Z_{p^{\infty}}$ is an artinian $\mathbb Z$-module.
1
vote
1answer
61 views

Is $\mathbb{Z}$ isomorphic to a direct subproduct of the family $\left\{ \mathbb{Z}_{n}\right\} _{n>1}$?

Is $\mathbb{Z}$ isomorphic to a direct subproduct of the family $\left\{ \mathbb{Z}_{n}\right\} _{n>1}$?
5
votes
2answers
168 views

$p$ prime, $P = \left\{ \frac{m}{p^e} \middle| m, e\in \mathbb{Z} \right\}$. Prove that $\mbox{Ext}(P; \mathbb{Z}) \cong \mathbb{Z}^{(p)}/\mathbb{Z}$

I don't know why the book Homology by Saunders Mac Lane is wwaaayyy tttoooo hard to digest. :((( This is like the third time I read this book, but still not clear is everything, and to tell the ...
2
votes
1answer
63 views

Modules decomposition into indecomposables

I think it's not true that every module (over arbitrary ring) is the sum of indecomposable modules, but I can't find counterexample and literature about this problem. Can anyone help me? Also I have ...
1
vote
2answers
372 views

Finding invariant factors of finitely generated Abelian group

There is this question that I wasn't sure how to do but somehow got the answers partially correct (maybe). Suppose that the abelian group $M$ is generated by three elements $x,y,z$ subject to the ...
4
votes
1answer
102 views

Intersection of direct summands of torsion free abelian groups

Suppose that $G_1$ and $G_2$ are direct summands of torsion free abelian group $G$. Must $G_1 \cap G_2$ also be a direct summand? It is true when $G$ is free, at least.
0
votes
1answer
82 views

$G/G' \cong I/I^2$ where $I$ is the augmentation ideal [duplicate]

Possible Duplicate: Isomorphism between $I_G/I_G^2$ and $G/G'$ Let $G$ be a finite group. Let $I\unlhd\mathbb{Z}[G]$ be the augmentation ideal of the integral group ring $\mathbb{Z}[G]$. ...
4
votes
1answer
150 views

Smallest pure subgroup containing a fixed subgroup

I will ask a slightly more precise question then in the title. Let $G$ be a finite abelian group, and $g_1, \ldots, g_n \in G$ such that the cyclic groups they generate are in direct sum $\langle g_1 ...
3
votes
3answers
237 views

Splitting exact sequences of finite abelian groups

I would like to find a condition for an exact sequence of abelian groups $$ 0\to H\to G\to K\to 0 $$ to split. Assume for simplicity that $H=\langle h \rangle$ is cyclic, and choose a basis for $G= ...
3
votes
3answers
165 views

Constructing a basis for finite abelian groups

Let $G$ be a finite abelian group, and $g_1, \ldots, g_k$ a set of "linearly independent elements", namely such that $\langle g_1 \rangle \oplus \ldots \oplus\langle g_k \rangle$. I would like to ...
3
votes
1answer
154 views

Linear algebra of finite abelian groups

Let $\phi:G \to H$ be a surjective homomorphism of finite abelian groups, and let $g_1, \ldots, g_n$ be an irredundant set of generators (from now on, a basis) for $G$. be a basis for $G$, meaning a ...
2
votes
2answers
302 views

Classify finitely generated modules over the ring $\mathbb{C}[\epsilon]$ where $\epsilon^2=0$

Classify finitely generated modules over the ring $\mathbb{C}[\epsilon]$ where $\epsilon^2=0$ Since $\mathbb{C}[x]$ s noetherian we have that $\mathbb{C}[x]/(x^2)$ is too. And thus finitely generated ...
0
votes
1answer
108 views

Finding subgroups of index 2 of $G = \prod\limits_{i=1}^\infty \mathbb{Z}_n$

I looked at this question and its answer. The answer uses the fact that every vector space has a basis, so there are uncountable subgroups of index 2 if $n=p$ where $p$ is prime. Are there ...
1
vote
1answer
234 views

Torsion module and its socle

A torsion abelian group has nonzero socle and is an essential extension of it. Let $R$ be a commutative ring. If $M$ is a unital, torsion $R$-module with nonzero socle, is $M$ an essential extension ...
1
vote
0answers
93 views

Computing a generating set of the kernel of a module

Given a generating set of a $\mathbb{Z}$-module $M \subseteq {\mathbb{Z}_k}^n$, is there a known algorithm to compute a generating set of $\{u \in {\mathbb{Z}_k}^n \, : \, \forall v \in M \quad v ...
6
votes
2answers
318 views

Whence this generalization of linear (in)dependence?

I recently came across a definition of (in)dependence that is supposed to be a generalization of linear (in)dependence among a set of vectors: An element $x$ is dependent on a set of elements ...
3
votes
1answer
345 views

Calculating Hom(A,B)

I have been studying modules and homological algebra as of late but somehow I have missed how to calculate Hom(A,B) for abelian groups, modules and Hom(A,_)/Hom(_,B) for exact sequences. I have no ...
7
votes
2answers
536 views

Can the tensor product of two non-free abelian groups be non-zero free?

It's pretty easy to construct an (R-S) bi-module M and a left S-module N such that neither M_S nor N is a projective S-module, but the tensor product is a non-zero projective R-module. However, ...